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Calculus conservation of momentum problem

  1. Dec 4, 2009 #1
    Hello All,

    I am trying to understand the solution of a problem that uses calculus and the conservation of momentum principle.


    The problem is here: http://www.physics.harvard.edu/academics/undergrad/probweek/prob79.pdf [Broken] and the answer is here: http://www.physics.harvard.edu/academics/undergrad/probweek/sol79.pdf [Broken]

    I have several questions concerning the solution, and I will ask one question at a time. My understanding of integration is weak. In the solution, we arrive at:
    [tex]\int_{M}^{m} \frac{dm}{m} = \int_{0}^{v} \frac{dv}{u-v}[/tex]. Why is it that we integrate between m and M and v and 0? What does integrating those terms physically mean? I think there is something missing in my understanding.
     
    Last edited by a moderator: May 4, 2017
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  3. Dec 5, 2009 #2

    Redbelly98

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    The limits of integration represent values taken initially (lower limit) and some time t later (upper limit).

    Initially, the mass is M (car only, no balls) and the velocity is 0. Some time later, these quantities are m and v. Or put another way, the mass is M when the velocity is 0, and the mass is m when the velocity is v.

    p.s.:
    By the way, whoever wrote that solution was sloppy by making variables in the integrand the same as the limits, which can be considered as constants for the purposes of integration. For example, it should read

    [tex]\int_{M}^{m} \frac{dm'}{m'} = \int_{0}^{v} \frac{dv'}{u-v'}[/tex]​

    where the primes (') are used to distinguish the variables of integration.
     
  4. Dec 5, 2009 #3
    Redbelly98,

    so m = mass of the car M plus the culmination of balls thrown inside the car at a later time?
     
    Last edited: Dec 5, 2009
  5. Dec 5, 2009 #4

    Redbelly98

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    Yes.

    (This is implied by equation (1) in the solution you linked to.)
     
  6. Dec 5, 2009 #5
    To further clarify things, m and v are functions and dm is the mass of the ball and dv is the change in the velocity of the car after being hit by the ball.

    Another question, why is [tex]dm/dt[/tex] = [tex] (u-v)\sigma/u[/tex]? I thought [tex]dm/dt[/tex] is the rate at which the ball hits the car, which is defined as [tex]\sigma[/tex].
     
  7. Dec 5, 2009 #6

    Redbelly98

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    σ is the rate at which the balls leave you (the person throwing them).

    The rate at which the balls hit the moving car is different. For example, if the car moves at the same speed as the balls (v=u), the balls do not hit the car and the rate of mass increase must be zero in that case.
     
  8. Dec 5, 2009 #7
    ok thanks for clarifying that issue. Sorry I'm at little slow, why is [tex]dm/dt[/tex] = [tex](u-v)\sigma/u[/tex]? Also, not sure how eqn (3) leads to (4), even though it is derived in the solution.
     
  9. Dec 6, 2009 #8

    Redbelly98

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    One would have to work out the rate at which mass arrives at the car, given u, v, and σ. If you assumed a single mass equal to σ·1s were thrown every second, with a speed of u, you could work out how often the masses would hit a car traveling at speed v (assumed constant for this calculation).

    Start with eqn. (2), and differentiate w.r.t. t. That gives an expression for dm/dt, which may be equated with the R.H.S. of (3). Separate the v and t parts of the equation, and integrate.
     
  10. Dec 6, 2009 #9
    I understand how (2) leads to (3) leads to (4). However, I'm stuck still on the rate at which the balls hit car. It didn't even occur to me that the balls hitting the car would be different from balls leaving the thrower until you told me.
     
  11. Dec 7, 2009 #10

    Redbelly98

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    Yes, that is probably a common error when looking at problems like this. If you are familiar with the Doppler shift already, this is an analogous effect.

    To help you think about it:

    Suppose the car is traveling at speed v. When it is a distance L from the thrower, a ball of mass σ·Δt is thrown toward the car at velocity u.

    Question #1: at what time t1 does this ball hit the car?

    Next: at a time Δt later, another ball of mass σ·Δt is thrown at the car.
    (Note: throwing a mass of σ·Δt at intervals spaced by Δt is consistent with the a mass-per-time rate of σ)
    When the second ball is thrown, the car is at a distance L+v·Δt from the thrower.

    Question #2: at what time t2 does this second ball hit the thrower?

    Question #3: what is t2 - t1?

    Note that a mass of σ·Δt hits the car every t2-t1 time interval, so the rate at which mass hits the car is

    σ·Δt / (t2-t1)​

    Use the answer to Question #3 to replace (t2-t1) in this expression, and find the rate at which mass hits the car.
     
  12. Dec 10, 2009 #11
    Sorry I can't get t2. Tried really hard.
     
  13. Dec 10, 2009 #12

    Redbelly98

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    Okay, here's a starter-hint:

    The second ball is thrown at time Δt. At this point, the car is a distance L+v·Δt from the thrower.

    How long does this baseball take to reach the car? Use the same method you used to find t1, using the new starting distance L+v·Δt instead of just L.

    Then add that time to Δt to get t2.
     
  14. Dec 10, 2009 #13
    I got a ridiculous answer, but I will share it with you anyway.

    Since distance = speed x time, L = initial distance between car and thrower, u = speed of ball, v1 = speed of car before being hit, v2 speed of car after being hit, t1 = time it takes for the first ball to hit, t2 = time it takes for the second ball to hit, Δt = time between first and second balls thrown.

    [tex]\because[/tex] t1(u-v1) = L+v1t1
    [tex]\therefore[/tex]t1 = [tex]L/(u-2v)[/tex]

    The distance it takes the ball to catch up with the car is equal to the initial distance L plus how much the car travels during the time it takes the ball to hit the car.

    (t1-Δt)(u-v1)+(t2-t1)(u-v2) = L+vt1+(t2-t1)v2
    This is the equation of distance that the car travels before being hit by the second ball. The speed is different after the car being hit by the first ball since it has an additional mass.
     
    Last edited: Dec 10, 2009
  15. Dec 11, 2009 #14

    Redbelly98

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    Hi,

    I haven't forgotten this, just haven't had time to make a good reply. Will try to respond tomorrow (Saturday).
     
  16. Dec 12, 2009 #15

    Redbelly98

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    That's not a bad start, I caught a couple of errors that I'll point out:
    The left-hand-side (LHS) here, the distance the 1st ball travels, is simply u·t1. That's because the ball is travelling at a speed u for a time t1. That will change your result for t1:
    Same error here; the 2nd snowball is traveling at speed u for a total time t2-Δt, so it's distance traveled is u·(t2-Δt).

    Hint: on the RHS, the term L+v1t1 may be replaced with u·t1.
     
  17. Dec 12, 2009 #16
    OK so here are the corrections.

    [tex]\because[/tex] u·t1 = L+v1t1
    [tex]\therefore[/tex] t1 = (L+ v·t1)/u

    I'm still having trouble with the second equation. This is what I got.

    u·(t2-Δt) = u·t1+(t2-t1)v2

    The last term of this equation deals with the motion of the car after being hit by the first snow ball until being hit by the second snowball. Then we have to calculate v2, but I'll leave that for later.
     
  18. Dec 12, 2009 #17

    Redbelly98

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    Looking good so far :smile:, and I agree, leave calculation of v2 for later.

    This last equation can be manipulated to determine "t2-t1" in terms of the other parameters.
     
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